Molcas Forum

rishu

Member

Registered: 2020-10-03

Posts: 25

difference contributions in the oscillator strengths

Hi
I want to compute different contributions in the oscillator strengths from RASSI i.e., electric-dipole-electric-dipole, electric-quadrupole-electric-quadrupole, magnetic-dipole-magnetic-dipole, electric-dipole-electric-octupole, electric-dipole-magnetic-quadrupole.
It will be helpful if someone will explain how to read that in the output as well.

Does the QIALL keyword print all the quadrupole moments, electric as well as magnetic?
I don't understand how to get their individual contribution.

Ignacio

Administrator

From: Uppsala

Registered: 2015-11-03

Posts: 1,097

Re: difference contributions in the oscillator strengths

QIALL should print all, but you also need to make sure you have included all required integrals (OMQI, MULT = 3), and possibly that all have the same origin (it should be the case by default, but in the case of charged systems, the dipole is not invariant and uses a different center). Example:

&GATEWAY
  Coord = 4
    angstrom
    C   0.00000000   0.00000159   0.02308959
    S  -0.00000000  -0.00000053   1.66405025
    H   0.93594257  -0.00000053  -0.56306999
    H  -0.93594257  -0.00000053  -0.56306999
  Basis = ANO-RCC-MB
  Group = NoSym
  RICD
  ANGM = 0.0 0.0 0.0
  OMQI = 0.0 0.0 0.0

&SEWARD
  Multipoles = 3
  Center = 4
    0  0.0 0.0 0.0
    1  0.0 0.0 0.0
    2  0.0 0.0 0.0
    3  0.0 0.0 0.0

&RASSCF
  nActEl=4
  Charge = 0
  SPIN = 1
  RAS2 = 6
  StAverage = 3

&RASSI
  EJOB
  QIAll
  QIPrint = 0.0

In the output, you'll have:
1a. Dipole transition strengths
1b. Velocity transition strengths
2. Magnetic-Dipole - Magnetic-Dipole transition strengths
3. Quadrupole transition strengths
4. Electric-Dipole - Electric-Octupole transition strengths
5. Electric-Dipole - Magnetic-Quadrupole transition strengths

For convenience, you also have
6. Second-order contribution to the transition strengths (sum of 2-5)

So (if I'm not mistaken), the final strengths would be 1b+6. You could take 1a+6, but then you'd be mixing different gauges and the result wouldn't be origin-independent (see https://doi.org/10.1016/j.cplett.2017.05.003).